Uniquely defined

Let n be a natural number. L.H.S and R.H.S will be natural number. Therefore n should be a square number. Let √n = x then the equation can be rewritten as

$x^{2}$ != $x^{4}$ + $x^{3}$

$x^{2}$ $(x^{2}-1$ )!= $x^{3}$ ( $x+1$ )

$(x+1)(x-1)$ $(x^{2}-2)$ != $x(x+1)$

$(x^2-2)! = \frac x {x-1}$

As $x$ is a natural number L.H.S and R.H.S will be natural number.(except at x=1 as 1/0 is not defined.)

x/x-1will only be natural when x = 2 therefore n= 4.

Pure algebra can be applied. Manipulate the formula to read as follows: ((n!-n^2)^(⅔))=n. Obviously n!-n^2 must equate to some x in the set Z+ such that x>3 because then a cube root can be taken. Upon testing numbers, we only haven to test 4, 5, and 6. We see that x=4 satisfies the equation. Upon analyzing the results for x=5 and x=6, it is clear that any x>4 cannot satisfy the equation. Thus x=4 is our only solution, or rather n=4.